3
$\begingroup$

I am reading through the dynamical systems theory and there is an example of a Mass-Spring system. The state equations are given by

$\displaystyle \frac{d x_1}{dx}(t) = x_2(t)$

$\displaystyle \frac{d x_2}{dx}(t) = \frac{-k}{m}x_1(t)+\frac{f(t)}{m}$

Then we find the equilibrium point by setting

$\displaystyle 0 = x_2(t)$

$\displaystyle 0 = \frac{-k}{m}x_1(t)+\frac{f(t)}{m}$

Which gives as result

$x_{eq} = \left[ \begin{array}{c} x_{1} \\ x_{2} \\ \end{array} \right] = \left[ \begin{array}{c} f/k \\ 0 \\ \end{array} \right]$

Finally, the book defines the increment with respect to the equilibrium point $\Delta x(t) = x(t) - x_{eq}$. Substracting equation (1,3) and (2,4) the result is

$\displaystyle \frac{d \Delta x_1}{dx}(t) = \Delta x_2(t)$

$\displaystyle \frac{d \Delta x_2}{dx}(t) = \frac{-k}{m}\Delta x_1(t)$

So far so good, but for the next step they get "the general solution, parametrized by the initial state" as

$\displaystyle \Delta x_1(t) = \Delta x_1(0) cos(\omega t) + \frac{\Delta x_2(0)}{\omega} sin(\omega t)$

$\displaystyle \Delta x_2(t) = -\Delta x_1(0) \omega sin(\omega t) + \Delta x_2(0)cos(\omega t)$

This result I don't understand where does it comes from, could someone give me a hint?

$\endgroup$
1
$\begingroup$

This is ODE business, the solution of : \begin{eqnarray*} \frac{dx_1}{dt} &=& x_2\\ \frac{dx_2}{dt} &=& -\frac{k}{m} x_1 \end{eqnarray*} is of the form you wrote. Let $\omega^2 = \frac{k}{m}$ for simplification.

A way to prove it is to differentiate the first equation and substitute the second in: $$\frac{d^2x_1}{dt^2} = \frac{dx_2}{dt} = -\omega^2 x_1$$

So the solution for $x_1$ is of the form : $$x_1(t) = A_1\cos(\sqrt{\omega}t) + A_2\sin(\sqrt{\omega}t)$$

Differentiate it to find the solution to $x_2(t)$.

Then substituting the initial conditions you find $A_1 = x_1(0)$ and $A_2 = \frac{x_2(0)}{\omega}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.